We usually neglect the mass of a spring if it is small compare...

Question

We usually neglect the mass of a spring if it is small compared to the mass attached to it. But in some applications, the mass of the spring must be taken into account. Consider a spring of unstretched length ℓ and mass M_S uniformly distributed along the length of the spring. A mass m is attached to the end of the spring. One end of the spring is fixed and the mass m is allowed to vibrate horizontally without friction (Fig. 7–31). Each point on the spring moves with a velocity proportional to the distance from that point to the fixed end. For example, if the mass on the end moves with speed v₀, the midpoint of the spring moves with speed v₀ / 2. <IMAGE>

Show that the kinetic energy of the mass plus spring when the mass m is moving with velocity v is

K = (1/2)Mv²

where M = m + (1/3)M_S is the “effective mass” of the system. [Hint: Let D be the total length of the stretched spring. Then the velocity of an infinitesimal length dx of spring, of mass dM, located at x is v(x) = v₀(x/D). Note also that dM = dx( M_S/D) .]

Show that the kinetic energy of the mass plus spring when the mass m is moving with velocity v is

K = (1/2)Mv²

where M = m + (1/3)M_S is the “effective mass” of the system. [Hint: Let D be the total length of the stretched spring. Then the velocity of an infinitesimal length dx of spring, of mass dM, located at x is v(x) = v₀(x/D). Note also that dM = dx( M_S/D) .]

Accepted Answer

Welcome back. Everyone in this problem. Seismic isolators used in earthquake prone areas consist of spring supporting a heavy base to dampen ground vibrations picture an isolator with a spring mass, MS natural length L and a heavy base mass M designed for horizontal frictionless movement. The spring's design ensures that the movement speed at each point is proportional to its distance from the fixed end during seismic activities. Determine the expression for the kinetic energy ke of this system when the base moves at speed. V for our answer choices. A says that it's half of a third of MS M all multiplied by V squared B says it's a half of the expression MS M all multiplied by V squared. C says it's a half multiplied by the expression two thirds of MS plus M multiplied by V squared. And D says it's a half of a half of MS plus M multiplied by V squared. Now, what are we trying to figure out here? Well, we are trying to find an expression for the kinetic energy for our system in terms of MS, that is the spring's mass in terms of M, the mass of the heavy base. And in terms of V that is when the base moves at speed V. So first let's ask ourselves, what do we know that relates kinetic energy to the variables we are trying to express it in? Well, recall, OK, that kinetic energy is defined as being equal to half of MV squared. OK. And if we're going to find the kinetic energy of this system, then the total kinetic energy it's called that ke total is going to be equal to the kinetic energy of our spring. OK. Plus the kinetic energy of our base. Now, for the kinetic energy of our base, OK. We already know that that kinetic energy is going to be equal to half of MV squared. We know the mass of our base is M and our base moves at a speed V. But now we need to find the kinetic energy for our spring. What can help us to do that? Well, the spring follows Hook's law which states that the force exerted by the spring is directly proportional to the displacement from its equilibrium position. That is its natural length. L. So if we think about our spring, it has a linear mass density. Let me put that in red here. OK? For our spring, OK. We know, let me come down a bit. So for this spring, we know that it has a linear mass density of lambda where lambda is equal to the mass of the spring. MS divided by the spring's length. Now on our diagram, let's consider a small element DX of the spring at a distance X from the fixed N. As you can see here, its velocity is going to be X divided by L all multiplied by V due to the linear velocity distribution. OK. So here, what that tells us then is that given that we have, let me make a note of that here, given that its velocity is X divided by L all multiplied by V. Then the kinetic energy of this element or the change in kinetic energy DK is going to be equal to half of Lambda multiplied by the X multiplied by its velocity X divided by L multiplied by V all squared. OK. Now, since we have that in mind here, if we can integrate or change in kinetic energy over the length of the spring, that will give us the spring's total kinetic energy. OK. In other words, we can say then that the kinetic energy of the spring is going to be equal to the integral of the change in kinetic energy. OK. And that means based on our formula that we have here, we're going to be integrating a half of lambda. OK. I should have written Lambda that way, a half of Lambda multiplied by XV, divided by L all squared. OK. DX. Because we are integrating our kinetic energy with respect to X. Those small elements are those small strips of DX. Now here, if we want to go any further, we can isolate our variables, let's come down to this line. OK? Because now we can take the constant sort of for expression here, which, well, let me, let me go ahead and square it first. OK. So now we're talking about a half of lambda multiplied by X squared V squared divided by L squared DX. OK. And now since we are integrating with respect to X, we can get rid of our constants which in this case would be a half of lambda multiplied by V squared divided by L OK. But again, sorry, divided by L squared. My apologies divided by uh L squared here. But again, OK, we've already defined lambda earlier as MS divided by L. So we can replace it here with MS divided by L. So let me write the rest of the integral here. So that's gonna be X squared DX. OK. And no, that means we're gonna have a half of MS divided by L multiplied by V squared divided by L squared multiplied by the integral of X squared DX between its limits. That is from the start of the spring to its final value. L because L is the length of the spring and we can even simplify this some more because L multiplied by elsewhere in our denominator is going to give us LQ. OK. So we have all of that multiplying the integral of X squared with respect to X between the bonds of zero and L. So now let's integrate when we integrate X squared, OK, we end up with X cubed divided by three. So really, then our answer is going to be a half of MS V squared divided by L cubed multiplied by X cubed divided by three between the boundaries of L and zero. Now, if we substitute L into our expression here, so that's gonna be a half really a half of MS V squared divided by L cubed multiplied by L cubed divided by three minus zero, cubed divided by three substitute or bonds here and subtract but no zero, cube divided by three is zero. So what we're really saying is that we're multiplying this entire thing by L cubed divided by three. OK. So I think it's fair here. I think I can get rid of what we have here and just leave it as L cubed divided by three. No L cube cancels L cubed. And if we, if we write out, OK, the kinetic energy of our spring, we're going to get a half multiplied by a third, which is 1/6 OK? 1/6 multiplied by MS V squared. OK. So this is the spring's total kinetic energy. Now that we have the springs total kinetic energy, we can add it to the kinetic energy of the heavy base. We already know that that was half MV squared. So solving then for our total kinetic energy OK. Therefore, that means the total kinetic energy is going to be equal to kinetic energy of the spring 1/6 of MS V squared plus half MV squared. OK. And now we notice that we can factor out a half and V squared from both terms. So when you factor at a half, we're left with a third of MS plus M all multiplied by V squared. So this is the total kinetic energy of the system. If we look back at our answer choices, that means a is the correct answer. Thanks for watching everyone. I hope this video helped.

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